Question

14. find the area of a quadrilateral ABCD in which AB=3 cm, BC = 4
cm, CD= 4 cm, DA= 5 cm and AC = 5 cm.
15 In cricket​

Answer 1

2S=5+4+3 ⇒ S=6

Area of ΔABC

$= \sqrt{s(s−a)(s−b)(s−c)}$

$= \sqrt{15(15−9)(15−8)(15−13)}$

$=\sqrt{6(1)(2)(3)} = \sqrt{36} = 6$

⇒For area of ΔADC,

⇒2S=5+4+5

⇒ S=7cm

⇒Area of ΔADC

$= \sqrt{7(7−5)(7−5)(7−4)}$

$= \sqrt{7×2×2×3}$

$=2 \sqrt{2}1=9.16 {m}^{2}$

⇒Area of quad ABCD= Area of (△ABC+△ADC)

=6+9.16

$=15.16 {cm}^{2} ≈15.2 {cm}^{2}$